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Andrei [34K]
3 years ago
15

The cruise control function on Georgina's car should keep the speed of the car within 3 miles per hour of the set speed. The set

speed is 55 miles per hour. Write a compound inequality to show the leaves that are within each range. Then graph the solutions.

Mathematics
2 answers:
motikmotik3 years ago
8 0
This should help. Sorry I had to make this rn.

ehidna [41]3 years ago
6 0

Answer:

Given,

Set speed of car = 55,

∵ The speed of the car within 3 miles per hour of the set speed.

So, the maximum speed of car = 55 + 3,

And, the minimum speed of car = 55 - 3

Let x shows the current speed of car,

x ≤ 55 + 3 or x ≥ 55 - 3

x - 55 ≤ 3 or x - 55 ≥ - 3

x - 55 ≤ 3 or -(x-55) ≤ 3

⇒ | x-55 | ≤ 3

Which is the required compound inequality,

Now, 55 - 3 ≤ x ≤ 55 + 3

52 ≤ x ≤ 58

⇒ [52, 58]

Which is the solution of the above compound inequality.

Also, the graph is shown below.

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There are no like terms.

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A store sells two brands of hand lotion. Brand X costs $3.25 for 5 fluid ounces.
olga_2 [115]

Answer:

\frac{3,25}{5}=0,65 \\ for brand x for 1 fluid ounces

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0,75-0,65=0,1

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Step-by-step explanation:

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3 years ago
For the rational expression 2x^2-7x-15/x^2-25, do both of the following,
Mrrafil [7]

Answer:

a. \frac{2x + 3}{x+5}

b. x ≠ -5 (Vertical asymptote) and x ≠ 5 (Hole)

Step-by-step explanation:

Factor the numerator (Grouping):

2x^2 - 7x - 15              

Two numbers that multiply to -30 and add to -7 = -3 and 10

[2x^2 - 10x] + [3x - 15]

(2x + 3)(x - 5)

Factor the denominator (Difference of Two Squares):

x^2 - 25 = (x +5)(x-5)

Factored Expression:

(x - 5) can be factored out of top and bottom as a hole-

\frac{(2x + 3)(x - 5)}{(x + 5)(x - 5)} = \frac{2x + 3}{x+5}

Variable Restrictions:

Denominator ≠ 0

x + 5 = 0\\x = -5

Vertical asymptote at x = -5 ⇒ x ≠ -5

3 0
3 years ago
I would like to know the value of b and c
Ratling [72]

Answer:

a = 3, b = 8, c = 14

Step-by-step explanation:

First, we can organize this, putting lower values first.

2, 3, 3, 5, 9, 9, 11, 12

a , b , c somewhere there

Given this information, one thing we can look at first is the mode. The mode is 3, so there must be more 3s than any other number. Currently, there are 2 3s and 2 9s, so there must be at least one more 3 and no more 9s to make that true. Therefore, a, b, or c is 3. Therefore, we have

2, 3, 3, 3, 5, 9, 9, 11, 12

2 of a, b, c somewhere in there

Next, the median is 8. In our current state, the median is 5. There are 9 numbers, with 11 total including the 2 remaining values. Because there will be an odd amount of values, the median must be a number on the list. Therefore, our list is

2, 3, 3, 3, 5, 8, 9, 9, 11, 12

1 of a, b, c somewhere in there

There are 4 numbers above the median and 5 numbers below right now. To balance this out, there must be another number above the median. As a consequence, the remaining value must be greater than 8.

Finally, we know that the range is 12, so maximum - minimum = 12. Because the remaining number must be greater than 8, the minimum number is 2, no matter what. Therefore,

maximum - 2 = 12

add 2 to both sides to isolate maximum

maximum = 14

There is no 14 currently on the list, so the remaining value must be 14.

Our a, b, and c are as follows, in order from smallest to largest:

3, 8, 14

8 0
2 years ago
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